\section{Subgroups}

\problem{}<s2s3share>
What elements do $S_2$ and $S_3$ share?
\vspace{2cm}



Consider the sets $\{1, 2\}$ and $\{1,2,3\}$. Clearly, $\{1, 2\} \subset \{1, 2, 3\}$. \par
Can we say something similar about $S_2$ and $S_3$?

\vspace{2mm}

Looking at \ref{s2s3share}, we may want to say that $S_2 \subset S_3$ since every element of $S_2$ is in $S_3$. \par
This however, isn't as interesting as it could be. Remember that $S_2$ and $S_3$ are \textit{groups}, not \textit{sets}: \par
their elements come with structure, which the \say{subset} relation does not capture.

\vspace{2mm}

To account for this, we'll define a similar relation: subgroups.

\definition{}
Let $G$ and $G'$ be groups. We say $G'$ is a \textit{subgroup} of $G$ (and write $G' \subset G$) if the following are true:\par
(Note that $x, y$ are elements of $G$, and $xy$ is multiplication in $G$)
\begin{itemize}
	\item the set of elements in $G'$ is a subset of the set of elements in $G$.
	\item the identity of $G$ is in $G'$
	\item $x,y \in G' \implies xy \in G'$
	\item $x \in G' \implies x^{-1} \in G'$
\end{itemize}

The above definition may look faily scary, but the idea behind a subgroup is simple. \par
Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \par

\vspace{2mm}

Say we have a set of four elements and only look at the first three. \par
$S_3$ fully describes all the ways we can arrange those three elements:

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1a) at (0, 0.5) {1};
	\node (2a) at (1, 0.5) {2};
	\node (3a) at (2, 0.5) {3};
	\node (4a) at (3, 0.5) {4};

	\node (2b) at (0, -2) {2};
	\node (3b) at (1, -2) {3};
	\node (1b) at (2, -2) {1};
	\node (4b) at (3, -2) {4};

	\draw[line width = 0.3mm, ->, ogreen]
		(4a)
		-- ($(4a) + (0, -1)$)
		-- ($(4b) + (0,1)$)
		-- (4b);

	\line{1a}{1b}
	\line{2a}{2b}
	\line{3a}{3b}

	\node[fill=white,draw=oblue,line width=0.3mm] at (1, -0.75) {$S_3$};

\end{tikzpicture}
\end{center}


\problem{}
Show that $S_3$ is a subgroup of $S_4$.

\vfill
\pagebreak



\definition{}
Let $G$ and $H$ be groups. We say that $G$ and $H$ are \textit{isomorphic} (and write $A \simeq B$) \par
if there is a bijection $f: G \to H$ with the following properties:
\begin{itemize}
	\item $f(e_G) = e_H$, where $e_G$ is the identity in $G$
	\item $f(x^{-1}) = f(x)^{-1}$ for all $x$ in $G$
	\item $f(xy) = f(x)f(y)$ for all $x, y$ in $G$
\end{itemize}

Intuitively, you can think of isomorphism as a form of equivalence. \par
If two groups are isomorphic, they only differ by the names of their elements. \par
The function $f$ above tells us how to map one set of labels to the other.



\problem{}
Show that $\mathbb{Z}_7^\times$ and $\mathbb{Z}_9^\times$ are isomorphic.
\hint{
	Build a bijection with the above properties. \\
	Remember that a group is fully defined by its multiplication table.
}
\vfill


\problem{}
Show that $\mathbb{Z}_{10}^\times$ and $\mathbb{Z}_5^\times$, and $\mathbb{Z}_4$ are isomorphic.
\hint{
	Build a bijection with the above properties. \\
	Remember that a group is fully defined by its multiplication table.
}


\vfill

\problem{}
Show that isomorphism is transitive. \par
That is, if $A \simeq B$ and $B \simeq C$, then $A \simeq C$.

\vfill
\pagebreak


\problem{}<firstindex>
How many subgroups of $S_4$ are isomorphic to $S_3$? \par

\vfill




\problem{}
What are the orders of $S_3$ and $S_4$? \par
How is this related to \ref{firstindex}?

\begin{solution}
	$|S_4| = |S_3| \times [S_4 : S_3]$

	\vspace{2mm}

	This solution is written using index notation, \par
	but the class doesn't need to know what it means yet.
\end{solution}

\vfill

\problem{}
$S_4$ also has $S_2$ and the trivial group as subgroups. \par
How many instances of each does $S_4$ contain?

\vfill


\problem{}
$(\mathbb{Z}_4, +)$ is also a subgroup of $S_4$. Find it! \par
How many subgroups of $\mathbb{Z}_4$ are isomorphic to $S_4$?.

\begin{solution}
	A good hint is \say{look at generators.}

	\vspace{4mm}

	There are four instances of $\mathbb{Z}_4$ in $S_4$, each of which is generated by a 4-cycle of $S_n$. \par
	(i.e, the group generated by $(1234)$ is isomorphic to $\mathbb{Z}_4$)
\end{solution}


\vfill
\pagebreak
